This question comes from Petersen's Riemannian geometry section 4.2.3.the rotation symmetric metric:
Consider the rotation symmetric metric $g = dr^2 + \rho^2 ds_{n-1}^2 = dr^2 + g_r$ where $ds_{n-1}^2$ is the metric on the unit sphere, $\rho = \rho(r)$.Denote $g_r = \rho^2ds^2_{n-1}$
Define the Hessian be $2\text{Hess}\ r = L_{\partial_r}g$.Where $L$ is the Lie derivative.
Then we have the following result:
$$\begin{aligned}
2 \text { Hess } r &=L_{\partial_{r}} g_{r} \\
&=L_{\partial_{r}}\left(\rho^{2} d s_{n-1}^{2}\right) \\
&=\partial_{r}\left(\rho^{2}\right) d s_{n-1}^{2}+\rho^{2} L_{\partial_{r}}\left(d s_{n-1}^{2}\right) \\
&=2 \rho\left(\partial_{r} \rho\right) d s_{n-1}^{2} \\
&=2 \frac{\partial_{r} \rho}{\rho} g_{r} .
\end{aligned}$$
If we compute Lie Derivative of the Hessian,then we have :
$$\begin{aligned}
L_{\partial_{r}} \text { Hess } r &=L_{\partial_{r}}\left(\frac{\partial_{r} \rho}{\rho} g_{r}\right) \\
&=\partial_{r}\left(\frac{\partial_{r} \rho}{\rho}\right) g_{r}+\frac{\partial_{r} \rho}{\rho} L_{\partial_{r}}\left(g_{r}\right) \\
&=\frac{\left(\partial_{r}^{2} \rho\right) \rho-\left(\partial_{r} \rho\right)^{2}}{\rho^{2}} g_{r}+2\left(\frac{\partial_{r} \rho}{\rho}\right)^{2} g_{r} \\
&=\frac{\partial_{r}^{2} \rho}{\rho} g_{r}+\left(\frac{\partial_{r} \rho}{\rho}\right)^{2} g_{r} \\
&=\frac{\partial_{r}^{2} \rho}{\rho} g_{r}+\text { Hess }^{2} r
\end{aligned}$$
The question is why Hessian square :$$\text { Hess }^{2} r = \left(\frac{\partial_{r} \rho}{\rho}\right)^{2} g_{r}$$
I try to deduce it from definition $\text{Hess}^2\ r(X,Y) = g(\nabla_X \partial_r,\nabla_Y \partial_r)$ ,but it's a bit messy,maybe we can start by noticing that Hessian is proportional to $g_r$?
Best Answer
First note that $\nabla_X\partial_r$ perpendicular to $\partial_r$ ,since $L_X|\partial_r|^2_g = 0.$ which implies $g(\nabla_X\partial_r,\partial_r) = 0$
So By definition $$\text{Hess}^2(r)(X,Y) = g(\nabla_X\partial_r,\nabla_Y\partial_r) = \text{Hess}(r)(\nabla_X\partial_r,Y) = \frac{\partial_r\rho}{\rho}g_r(\nabla_X\partial_r,Y)$$.
Now comes the point,I claim $g_r(\nabla_X\partial_r,Y) = g(\nabla_X\partial_r,Y)$,this is because $dr^2(\nabla_X\partial_r,Y) = dr(\nabla_X\partial_r)dr(Y) = g(\partial_r,\nabla_X\partial_r)dr(Y) = 0$ since the perpendicular condition.
finally $$\frac{\partial_r\rho}{\rho}g_r(\nabla_X\partial_r,Y) = \frac{\partial_r\rho}{\rho}g(\nabla_X\partial_r,Y)=\frac{\partial_r\rho}{\rho}\text{Hess(r)}(X,Y) = \frac{\partial_r\rho}{\rho}\frac{\partial_r\rho}{\rho}g_r(X,Y) $$